Problem 101P

A 10.0-mL sample of a commercial hydrogen peroxide (H2O2) solution is titrated with 0.0998 M KMnO4. The end point is reached at a volume of 34.81 mL. Find the mass percent of H2O2 in the commercial hydrogen peroxide solution. (Assume a density of 1.00 g/mL for the hydrogen peroxide solution.) The unbalanced redox reaction that occurs in acidic solution during the titration is:

H2O2(aq) + MnO4− (aq) → O2(g) + Mn2+ (aq)

Solution:

Here we have to calculate the mass percent of H2O2 in sample.

Step 1:

The balanced equation for the redox reaction is

2 MnO−4 (aq) + 5H2O2 (aq) + 6 H+(aq) → 2 Mn2+(aq) + 5 O2(g) + 8 H2O(l)

Step-3

Molar mass of H2O2 = 34.014 g/mol

Mass percent = (grams of solute / grams of solute plus solvent) 100

Volume of H2O2 = 10.0 mL

Density of H2O2 = 1.00 g/mL

It is known that d = m/V

m = d V

= 1.00 g/mL10.0mL = 10 g (mass of H2O2)

Concentration of KMnO4 = 0.0998 M

Final volume = 34.81 mL= 0.3481 L

C = moles / liters

Or, C = (grams / Molar mass) / (liters)

Or, C = (m / M ) / V = m / (M V)

m = CMV

= 0.998 158.034 0.3481 = 54.901g

Hence total mass of the sample = 54.901 + 10 = 64.901 g

Mass percent of H2O2= =

= = 2.6204100= 262.04 %

Thus the mass percent of H2O2 in sample is 262.04%.